Selected option in dropdown from mysql/php reference

Question

Trying to get an option selected in a drop down based on the reference stored in a mysql database.

This is happening during producing a table already within a while loop.

I have tried the following focusing on the

<option selected='<? if ( $row2xsecid == $row1x[item_sub_sec_id]) { echo "selected"; } ?>' value="<? echo $row1x[item_sub_sec_id]; ?>"><? echo $row1x[item_sub_sec_title]; ?></option>

I cannot seem to get the option selected based on the reference stored in the mysql.

The snippet

  <div class="input-group">
                    <span class="input-group-addon">Select Group</span>
                    <select id='add_item_groupid' class="form-control">
                    <?
                    $sql1x = 'SELECT * FROM items_sub_section_list ORDER BY item_sub_sec_id ASC';
                    $result1x = mysql_query($sql1x);

                    while ($row1x = mysql_fetch_array($result1x)) {
                    ?>
                            <?
                            $sql2x = 'SELECT * FROM items_groups WHERE item_id=$itemid';
                            $result2x = mysql_query($sql2x);
                            while ($row2x = mysql_fetch_array($result2x)) { $row2xsecid = $row2x[item_sub_sec_id]; }
                            ?>
                        <option selected='<? if ( $row2xsecid == $row1x[item_sub_sec_id]) { echo "selected"; } ?>' value="<? echo $row1x[item_sub_sec_id]; ?>"><? echo $row1x[item_sub_sec_title]; ?></option>

                    <?
                    } 
                    ?> 
                    </select>
    </div>

The pastebin of the full script - http://pastebin.com/NZrETate

are u sure ur short tag is on <? in php.ini ?

Abhik Chakraborty
– Abhik Chakraborty

2014-03-27 16:34:02 +00:00
Commented Mar 27, 2014 at 16:34 — Abhik Chakraborty
– Abhik Chakraborty, Commented Mar 27, 2014 at 16:34

johan · Accepted Answer · 2016-05-16 15:38:50Z

I see you are fetching array's instead of associative arrays. (line 27: mysql_fetch_array instead of: mysql_fetch_assoc)

Doing this you have to keep in mind that the index of your fields start with 0 and are numerical which means this will return Undefined index:

line40: <td><? echo $row['item_id']; ?></td>

Because you fetched an array this will also return Undefined index: $itemid = $row['item_id']; Because $itemid is undefined the query at line 49 will also fail:

 $secid = mysql_query(" SELECT * FROM item_groups WHERE item_id='$itemid' ");

I also can't see where item_sub_sec_id is declared anywhere? (In PHP if this is a variable it must have a $ in front: $item_sub_sec_id)

Are you able to see the errors PHP is throwing you? If not I suggest showing all errors like so:

ini_set('display_errors', 1);
ini_set('display_startup_errors', 1);
error_reporting(E_ALL);

Once you have these issues sorted we can look at the problem, it will also be easier for me to give you a solution if we know what the DB structure looks like.

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